Problem: Divide the following complex numbers. $ \dfrac{1-8i}{2-3i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2+3i}$ $ \dfrac{1-8i}{2-3i} = \dfrac{1-8i}{2-3i} \cdot \dfrac{{2+3i}}{{2+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(1-8i) \cdot (2+3i)} {(2-3i) \cdot (2+3i)} = \dfrac{(1-8i) \cdot (2+3i)} {2^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(1-8i) \cdot (2+3i)} {(2)^2 - (-3i)^2} = $ $ \dfrac{(1-8i) \cdot (2+3i)} {4 + 9} = $ $ \dfrac{(1-8i) \cdot (2+3i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({1-8i}) \cdot ({2+3i})} {13} = $ $ \dfrac{{1} \cdot {2} + {-8} \cdot {2 i} + {1} \cdot {3 i} + {-8} \cdot {3 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{2 - 16i + 3i - 24 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{2 - 16i + 3i + 24} {13} = \dfrac{26 - 13i} {13} = 2-i $